On more bent functions from Dillon exponents

Transcription

1 AAECC DOI /s ORIGINAL PAPER On more bent functions from Dillon exponents Long Yu 1 Hongwei Liu 1 Dabin Zheng Receive: 14 April 014 / Revise: 14 March 015 / Accepte: 4 March 015 Springer-Verlag Berlin Heielberg 015 Abstract In this paper, we obtain several new classes of p-ary bent functions, where p is a prime. The bentness of all these functions is characterize by some exponential sums, which have close relations with Kloosterman sums. Moreover, we obtain some concise criterions on the bentness of p-ary functions in some special cases. In aition, our work generalizes some main results obtaine by Li et al. IEEE Trans Inf Theory 593): , 013). Keywors Dickson polynomial Kloosterman sum p-ary bent function 1 Introuction In 1976, Rothaus [13] introuce boolean bent functions which are maximally nonlinear boolean functions with even number of variables. It means that boolean bent functions achieve the maximal Hamming istance between boolean functions an affine functions. Since then, boolean bent functions have attracte much attention ue to their application in coing theory, cryptography an sequence esign. Later, Kumar et al. [8] generalize the notion of boolean bent functions to the case of functions over B Hongwei Liu hwliu@mail.ccnu.eu.cn Long Yu longyu@mails.ccnu.eu.cn Dabin Zheng zheng@hubu.eu.cn 1 School of Mathematics an Statistics, Central China Normal University, Wuhan , Hubei, China School of Mathematics an Statistics, Hubei University, Wuhan 43006, Hubei, China 13

2 L. Yu et al. an arbitrary finite fiel F p n, where p is a prime an n is a positive integer. Some results on constructions of bent functions on monomial, binomial an quaratic functions can be foun in [1 7,9,10,1,14 16]. Let p be a prime an m be a positive integer, n = m an F p n be the finite fiel with p n elements, an F p n = F p n \{0}. LetTrn 1 ) be the trace function from F p n to F p, i.e. x) = n 1 x pi for all x F p n. The bentness of boolean monomials with Dillon exponents was characterize by Dillon in [4] an Charpin et al. in []. The corrosponing p-ary case was investigate by Helleseth et al. [5]. Some multinomial bent functions with Dillon exponents were investigate in [7,1,14,15]. Recently, Li et al. [10] investigate the bentness of several special classes of p-ary functions of the following form f x) = p m 1 ) ) a i x ipm 1) + Tr o) 1 bx pn 1, 1.1) where a i F p n, b F p o), is a positive integer with p m + 1) an o) is the smallest positive integer satisfying o) n an p o) 1). The bentness of some special classes of p-ary functions with Dillon exponents pm ± l)p m 1) is etermine by some exponential sums, most of which have close relations with Kloosterman sums. This kin of Dillon exponents leas to some etaile new characterizations for the bentness of p-ary functions, an some new p-ary bent functions are obtaine. In this paper, we first investigate the bentness of several classes of p-ary functions of the form 1.1) with Dillon exponents i pm +l)p m 1), where i {0,..., 1}. This kin of Dillon exponents is a generalization of exponents pm ±l)p m 1) an it also leas us to etaile new characterizations for the bentness of p-ary functions. Following the section in [7] we further investigate a relationship between some partial exponential sums an Kloosterman sums. Base on this relationship, we get a new characterization for the bentness of a class of binomial p-ary functions see Theorem 3.15). Last, the bentness of other classes of bent functions with Dillon exponents of the form 1.1) is also characterize by some exponential sums. In some special cases, we can characterize the bentness of these functions from simple approaches see Corollaries 3.11, 3.1, 3.). The remainer of this paper is organize as follows. Section gives some preliminaries. In Sect. 3, the bentness of several classes of functions is characterize by some exponential sums. The concluing remarks are given in Sect. 4. Preliminaries Throughout this paper, we fix some notations. p is a prime, an m is a positive integer, n = m an is a ivisor of p m + 1. ω = e π 1 p is a complex primitive p-th root of unity, an α is a primitive element of F p n. 13

3 On more bent functions from Dillon exponents In the following, we give some basic efinitions an results. Definition.1 Let f : F p n F p be a p-ary function. The Walsh transform of f is efine by W f λ) = ω f x) Trn 1 λx),λ F p n. x F p n Definition. Let f : F p n F p be a p-ary function. Then f x) is calle a bent function if W f λ) = p n for all λ F p n.ap-ary bent function f x) is sai to be regular if for all λ F p n, W f λ) = p n ω f λ) for some function f from F p n to F p. The function f x) is calle the ual of f x). Remark.3 In particular, for p =, a boolean bent function is always regular. Definition.4 The Dickson polynomial D r x) F [x] of egree r is efine by ) k where = s D r x) = s 1 k j) sj=1 j r/ r r i ) r i x r i, r =, 3,..., i { an r r/, if r is even; = r 1)/, otherwise. Definition.5 Let β F p m, the Kloosterman sum K m β) over F p m is efine as K m β) = ω Trm 1 βx+x pm ). x F p m Let U ={x x pm = 1, x F p n } be a cyclic subgroup of F pn. It is easy to see that U can be ecompose into U = 1 k=0 V k, where V 0 ={ξ i 0 i < pm }, V k = ξ k V 0 for 1 k 1, an ξ is a generator of the cyclic group U. For i = 0, 1,..., 1 an a F p n, we efine S i a) = ω Trn 1 aξ i x). It is well known that if p =, then If p >, then 1 ω Trn 1 ax) = S i a) = 1 K m a), a F m. 1 ω Trn 1 ax) = S i a) = 1 K m a ) pm, a F p n, 13

4 L. Yu et al. which is given in [5]. In particular, if p = an = 3, Mesnager [11] gavea relationship between S i a) an some well-known exponential sums, an then obtaine a new class of binomial bent functions. Furthermore, if p = an = 5, S i a) was etermine by some well-known exponential sums in [14]. By using these results, Tang et al. [14] characterize the bentness of a new class of binomial functions. For p > an =, the only known results on S i a), where i = 0, 1, were obtaine by Jia et al. [7], which were use to characterize a new class of binomial bent functions. Following this line, the bentness of more p-ary functions can be concisely characterize if S i a) is obtaine for some p an, where 0 i 1. When p =, Li et al. [10] obtaine a relation between S 0 a) an some exponential sums as follows. Lemma.6 [10] Let p =, a= aξ k F n, where 0 k m, an a F m.if k 0 mo ), then S 0 a) = 1 + E m,a) K m a), where E m, a) = x F m 1)Trm 1 ad x)), ξ is a generator of the cyclic group U. Let p be an o prime, C t ={α i+t i = 0, 1,..., pn 3 For a F p n an b C 0, we efine Ra) an Qb) as follows: } F p Ra) = 1 K m a p m ), Qb) = Tr m 1 b pm ). When =, the values of S 0 a) an S 1 a) can be obtaine as follows. Lemma.7 [7] Let the notations be given as above. We have an where I = { Ra) + I ω S 0 a) = Qa) ω Qa) ), a C 0 + ; Ra), otherwise, { Ra) I ω S 1 a) = Qa) ω Qa) ), a C 0 + ; Ra), otherwise, 1) 3m p m 1) m p m, p 3 mo 4);, otherwise, n, where t = 0, 1. an C + 0 ={a C 0 Qa) = 0}. If p m 3 mo 4), = 4 an a = α j pm ) C 0, where j is a positive integer with 0 j p m, we get the following relationship between S i a) an Kloosterman sum, where i = 1, 3. 13

5 On more bent functions from Dillon exponents Proposition.8 Let the notations be given as above. We have S 1 a) = S 3 a) = Ra) I ωqa) ω Qa) ), where Ra), Qa), I are given in Lemma.7. Proof Note that p m 3 mo 4), wehave4 p m + 1), hence S 1 a) = ω Trn 1 aξ x) = ω Trn 1 aξ pm x pm ) = ω Trn 1 aξ 3 ξ pm 3 x) = ω Trn 1 aξ 3 x) = S 3 a). Since S 1 a) + S 3 a) = x H ωtrn 1 aξ x), where H ={ξ i Lemma.7, then 0 i pm 1 }, by S 1 a) = S 3 a) = { Ra) I ω Qa) ω Qa) ), a C 0 + ;, otherwise. Ra) We have Qa) = 0 when a C 0 \C 0 +, then the result follows. As we known, for an o prime p, everyx F pn has a unique representation as x = uy, where u U ={1,α,...,α pm }, y F pm. The following result can be easily obtaine. Proposition.9 For λ F p n, there exists only one solution in U such that Trn m λu) = 0. To investigate the bentness of f x) efine by 1.1), we nee the following lemma. Lemma.10 [5] Let p be an o prime, f : F p n F p be a regular bent function such that f x) = f x) an f 0) = 0, then f 0) = 0, where f is the ual function of f. A necessary an sufficient conition such that f x) efine by 1.1) is regular bent, which was given in [10]. We restate this result an give a ifferent proof. Lemma.11 [10] Assume the notations given as above. Then the function f x) efine by 1.1) is regular bent if an only if Sa 1, a,...,a p m 1, b) = 1, where Sa 1,...,a p m 1, b) = ω p m 1 a i x i )+Tr o) 1 bx pm ). 13

6 L. Yu et al. Proof We first compute the Walsh transform of f x). Ifλ = 0, then W f 0) = If λ F p n, then W f λ) = x F p n ω f x) = 1 + u U = 1 + p m 1 ) ω u U = 1 + p m 1 ) ω p m 1 p m 1 p m 1 ω y F p m a i u ipm 1) ) +Tr o) 1 bu pn ) 1 a i u ipm 1) ) +Tr o) 1 bu pn ) 1 ai x i ) +Tr o) 1 bx pm ) = 1 + p m 1 ) S a 1, a,...,a p m 1, b )..1) x F p n = 1 + u U = 1 + u U u U ω = 1 + u U ω ω = 1 + p m ω ω f x) Trn 1 λx) p m 1 ω y F p m y F p m p m 1 p m 1 p m 1 p m 1 ω p m 1 a i u ipm 1) ) +Tr o) 1 bu pn ) 1 λuy) a i u ipm 1) ) +Tr o) 1 bu pn ) 1 λuy) a i u ipm 1) ) +Tr o) 1 bu pn ) 1 a i u ipm 1) ) +Tr o) 1 bu pn ) 1 ai x i ) +Tr o) 1 bx pm ) a i u ipm ) 1) λ +Tr o) 1 buλ p n 1 ) y F p m ω Trm 1 ytr n m λu)) S a 1, a,...,a p m 1, b ),.) where u λ satisfies Tr n m λu) = 0 an the last equality in.) is obtaine by Proposition.9. Now, we finish the proof by two cases. 13

7 On more bent functions from Dillon exponents Case I, p =. If Sa 1, a,...,a m 1, b) = 1, we can get f x) is bent from Eqs..1) an.). Conversely, if f x) is bent, then W f 0) = 1 + m 1)Sa 1, a,...,a m 1, b) {± m }. Since Sa 1, a,...,a m 1, b) is an integer, then Sa 1, a,...,a m 1, b) = 1. Case II, p >. If f x) is regular bent, then W f 0) = p m ω f 0) by Definition.. By Lemma.10,wehaveW f 0) = 1+p m 1)Sa 1, a,...,a p m 1, b) = p m. Therefore, we get Sa 1, a,...,a p m 1, b) = 1. Conversely, if Sa 1, a,...,a p m 1, b) = 1, it is easy to fin that f x) is regular bent from Eqs..1) an.). 3Binaryanp-ary bent functions In this section, we stuy several classes of bent functions with Dillon exponents of the form 1.1) forp = an p >, respectively. 3.1 Binary bent functions In the following, we investigate two classes of bent functions of the form 1.1) First class of binary bent functions We assume that an l are positive integers such that gcl, m ) = 1, an stuy the bentness of the following functions 1 f a0,...,a 1,b x) = a i x l+i m ) ) m 1) + Tr o) 1 ) bx n 1, 3.1) where a i F n,0 i 1, b F o). In what follows, we give a general characterization on the bentness of function f a0,...,a 1,bx) efine by 3.1). Base on this characterization, we investigate two special classes of binary bent functions of the form 3.1). Moreover, from these two special classes of binary bent functions, we can obtain some binary bent functions, which ha been iscusse by Li et al. [10]. Proposition 3.1 Let the notations be given as above. Then the function f a0,...,a 1,bx) efine by 3.1) is bent if an only if 1 1) Tro) 1 bξ j m ) ) 1) 1 a i ξ j i m Proof By Lemma.11, f a0,...,a 1,bx) is bent if an only if ) ) ξ jl x = 1. 1) 1 a i x l+i m )+Tr o) 1 bx m ) = 1. 13

8 L. Yu et al. Note that 1) 1 = 1) a i x l+i m ) +Tr o) 1 bx m ) ) = 1) ) 1 = Tr n 1 ai x l) +Tr o) 1 b) + 1) 1 a i ξ l+i m x )+Tr l o) 1 bξ m ) ) ) 1 a i ξ 1) l+i m )x l +Tr o) 1 bξ 1)m ) a i x)+tr o) 1 b) + 1) 1 a i ξ i m ξ l x 1 a i ξ 1) i m ) ) ) ξ 1)l x +Tr o) 1 bξ 1)m ) ) Tr o) 1 bξ jm ) 1) 1) 1 a i ξ j i m ) ξ jl x ) ) +Tr o) 1 bξ m ), 3.) then we finish the proof. From Proposition 3.1, the necessary an sufficient conition on the bentness of f a0,...,a 1,bx) efine by 3.1) is inee complex. However, if we take some special values of a i an b, then the bentness of f a0,...,a 1,bx) efine by 3.1) can be concisely characterize as follows. Theorem 3. Let a 0 F m, a 1 = a = = a 1 F m f a0,...,a 1,0x) efine by 3.1) is bent if an only if an a 0 = a 1, then K m a 0 ) + 1)K m a 0 + a 1 ) { Em, a = 0 ) + 1)E m, a 0 + a 1 ) ), if l; E m, a 0 ) E m, a 0 + a 1 ) ), if gc, l) = 1, where E m, a) is given in Lemma.6. Proof Since ξ j m is a root of 1 + z + z + +z 1 = 0 for each 1 j 1, an a 1 = a = = a 1, we get 1 a iξ i j m ) ξ jl x) = Tr n 1 a0 + a 1 )ξ jl x ) for each 1 j 1. Note that b = 0 an gcl, m ) = 1, then Eq. 3.) is equivalent to 13

9 On more bent functions from Dillon exponents 1) 1 a i x l+i m ) = 1) Trn 1 a 0x) + 1) Trn 1 a 0+a 1 )ξ l x) + + 1) Trn 1 a 0+a 1 )ξ 1)l x). In the following, we iscuss Eq. 3.3) in two cases. 1. If l, by Lemma.6, then 1) 1 a i x l+i m ) = = 1 + E m,a 0 ) K m a 0 ) 3.3) 1) Trn 1 a0x) + 1) 1) Trn 1 a 0+a 1 )x) + 1) 1 + E m,a 0 + a 1 ) K m a 0 + a 1 ). Hence, by Proposition 3.1, f a0,...,a 1,bx) is bent if an only if K m a 0 ) + 1)K m a 0 + a 1 ) = E m, a 0 ) + 1)E m, a 0 + a 1 ) ).. If gc, l) = 1, we have {l mo ), l mo ),..., 1)l mo )} ={1,,..., 1}. By Lemma.6, 1) 1 a i x l+i m ) = 1) Trn 1 a 0x) + 1) Trn 1 a 0+a 1 )ξ x) + + 1) Trn 1 a 0+a 1 )ξ 1 x) = 1) Trn 1 a 0x) + x V 1 1) Trn 1 a 0+a 1 )x) + + = 1) Trn 1 a 0x) + x V 1 1) Trn 1) Trn 1 a 0+a 1 )x) 1) Trn 1 a 0+a 1 )x) = 1 K m a 0 + a 1 ) E m,a 0 ) K m a 0 ) 1 + E m,a 0 + a 1 ) K m a 0 + a 1 ). Therefore, by Proposition 3.1, f a0,...,a 1,bx) is bent if an only if K m a 0 ) + 1)K m a 0 + a 1 ) = E m, a 0 ) E m, a 0 + a 1 ). 1 a 0+a 1 )x) This finishes the proof. 13

10 L. Yu et al. Example 3.3 Let n = m = 6, = 9 an l = 1, a 0 F, a 3 1 = a = =a 8 F, 3 then f a0,...,a 8,0x) = Tr 6 1 a 0x 7 ) + 8 Tr 6 1 a 1x 71+i) ). By using Maple, there exist 9 pairs a 0, a 1 ) such that f a0,...,a 8,0x) is bent. If we take = 3 in Theorem 3., an combine the results on S i a) in [11], i = 0, 1,, we obtain the following result, which is exact [10, Corollary 1]. Corollary 3.4 [10] Let = 3, b= 0, a 0 F m, a 1 = a F m an a 0 = a 1, then f a0,a 1,a,0x) efine by 3.1) is bent if an only if K m a 0 ) + K m a 0 + a 1 ) = where C m a) = a F m 1)Trm 1 ax3 +ax). { Cm a 0 ) + C m a 0 + a 1 )), if3 l; C m a 0 ) C m a 0 + a 1 )), otherwise, For b = 0, by a similar proof as that in Theorem 3., we obtain the following result. Theorem 3.5 Let b = 0, l, a 0 F m, a 1 = a = =a 1 F m an a 0 = a 1, then f a0,...,a 1,bx) efine by 3.1) is bent if an only if ρk m a 0 ) + σ K m a 0 + a 1 ) = ρ E m, a 0 ) + σ E m, a 0 + a 1 )) + ρ + σ, where ρ = 1) Tro) 1 b), σ = 1 Lemma.6. j=1 1)Tro)bξ j m ) 1 an E m, a) is given in Example 3.6 Let n = m = 4, = 5, l = 5, an α be a primitive element of F 4, a 0 F, a 0 = a 1, a 1 = a = a 3 = a 4 F, b F, then f 4 a0,...,a 4,bx) efine by 3.1) is equal to Tr 4 1 a 0x 15 ) + 4 Tr 4 1 a 1x 35+i) ) + Tr 4 1 bx3 ). By using Maple, the number of a 0, a 1, b) such that f a0,...,a 4,bx) is bent is 60. The following result is a corollary of Theorem 3.5, which is exact [10, Theorem 3]. Corollary 3.7 [10] Let l, a 0 F m an a 1 = =a 1 = 0, then f a0,0,...,0,bx) efine by 3.1) is bent if an only if 1 1) Tro) 1 bξ j m ) = where E m, a) is given in Lemma E m, a 0 ) K m a 0 ), Remark 3.8 Some other special cases of Proposition 3.1 can be iscusse as above. Li et al. [10] investigate the bentness of binary functions with Dillon exponent m ± l) m 1), an then obtaine some new binomial, trinomial bent functions. In this subsection, we stuy the bentness of binary functions with Dillon exponent l + i m ) m 1), where i {0,..., 1}, an we give some new ifferent 13

11 On more bent functions from Dillon exponents characterizations on their bentness. Since the bentness of the function efine by 3.1) for the case of a = = a = 0 ha been iscusse in [10] an other cases are novel. From Theorem 3. an Theorem 3.5, we can obtain some new multinomial bent functions for suitable values of a i an b see Example 3.3 an Example 3.6) Secon class of binary bent functions In this subsection, we assume that s, k, r are integers an g = gcr, m + 1). Let f a,r,s x) = m g 1 ax ri+s)m 1) ), 3.4) where a F n an f 0) = 0. In the following, we give a necessary an sufficient conition on the bentness of f a,r,s x) efine by 3.4) for arbitrary r. Theorem 3.9 Let the notations be given as above. 1. If gcs, m + 1) = 1, 0 k m an a = aξ k F n with a F m, then f a,r,s x) efine by 3.4) is bent if an only if K m a) = g x g =1, 1) Trn 1 ax).. If gcs, m + 1) =, 0 k < m an a = aξ k F n with a F m, then f a,r,s x) efine by 3.4) is bent if an only if S 0 a) = x g =1, where S 0 a) is given in Lemma.6. 1) Trn 1 axs) + 1 g, Proof By Lemma.11, f a,r,s x) is bent if an only if 1) m g 1 axri+s ) = 1. Note that, m g 1 axri+s) ) = axs ) when x g = 1 an x U. Since m g 1 is even, then m g 1 1) axri+s) = g when x g = 1 an x U. So we have 13

12 L. Yu et al. 1) m g 1 axri+s ) = \x =1 1) Trn1axs) + g = g +g 1) Trn 1 axs). x g =1, 1) Trn 1 axs ) If gcs, m + 1) = 1, then 1)Trn 1 axs) = 1)Trn 1 ax) = 1 K m a) an gcs, g) = 1. Thus f a,r,s x) is bent if an only if K m a) = g x g =1, 1) Trn 1 ax). If gcs, m + 1) = an a = aξ k,wehave 1) Trn 1 axs) = 1) Trn 1 ax) = S 0 a). Hence, f a,r,s x) is bent if an only if S 0 a) = x g =1, 1) Trn 1 axs) + 1 g. Let s 1 be a positive integer, an g = gcr, m + 1) = 1, s = s 1 r.by3.4), we have f a,r,s1 r x) = m 1 axri+s 1) m 1) ). 3.5) Accoring to Theorem 3.9, we have the following result, which is exact [10, Theorem 4]. Corollary 3.10 [10] Let the notations be given as above. 1. If gcs 1 r, m + 1) = 1, 0 k m an a = aξ k F n with a F m, then f a,r,s1 r x) efine by 3.5) is bent if an only if K m a) = 1 1) Trn 1 a).. If gcs 1 r, m + 1) =, 0 k < m an a = aξ k F n with a F m, then f a,r,s1 r x) efine by 3.5) is bent if an only if where S 0 a) is given in Lemma S 0 a) = 1) Trn 1 a),

13 On more bent functions from Dillon exponents In particular, if g = 3, gcs, m ) = 1, then the following result can be obtaine from Theorem 3.9. Corollary 3.11 Assume the notations given as above. Let a = aξ k F n with a F m an 0 k m an f 0) = 0, then f a,r,s x) efine by 3.4) is bent if an only if K m a) = 3 Furthermore, if f a,r,s x) is bent, then 1) Trn 1 aξ j m 3 ). { K m a) = 0, if a) = Trn 1 aξ m 3 ) = aξ m 3 ) = 0; 4, otherwise. Proof By Theorem 3.9, f a,r,s x) efine by 3.4) is bent if an only if K m a) = 3 1) Trn 1 aξ j m 3 ). Since ξ m 3 + ξ m 3 = 1, then a) = Trn 1 aξ m 3 ) + aξ m 3 ). Itiseasy to check that K m a) = 0if a) = Trn 1 aξ m 3 ) = aξ m 3 ) = 0 an in other cases, K m a) = 4. This completes the proof. Example 3.1 Let α be a primitive element of F 6, n = m = 6, r = 3, s = 1, a F 6, then f a,3,1 x) = Tr 6 1 ax8 ) + Tr 6 1 ax49 ). By applying Maple, the number of this class of binomial bent functions is 36. Remark 3.13 Some other special cases of Theorem 3.9 can also be iscusse as above. Theorem 3.9 is a generalization of Theorem 4 in [10], an erive a concise characterization on the bentness of f a,r,s x) efine by 3.4) see Corollary 3.11). 3. p-ary bent functions In this subsection, let p be an o prime. The bentness of two classes of p-ary functions of the form 1.1) is characterize by some exponential sums, which have close relations with Kloosterman sums First class of p-ary bent functions Helleseth et al. [5] characterize the bentness of monomial Dillon functions f x) = axlpm 1) ) with gcl, p m + 1) = 1 by Kloosterman sum. Jia et al. in [7] stuie the bentness of binomial functions f x) = axlpm 1) )+bx pn 1 with gcl, p m + 13

14 L. Yu et al. 1) = 1. Later, Zheng et al. [15] further investigate the bentness of this class of binomial functions uner the case of gc l, pm + 1) = 1. In [16], Zheng et al. propose a class of binomial functions f a,b x) = ax lpm 1) ) + Tr 1 ) bx pn 1 4, 3.6) where p m 3 mo 4), a F p n, b F, an l is an integer with gcl, pm p 4 ) = 1, an etermine the bentness of these functions by subsequences of two sequences. In this subsection, we further investigate the bentness of f a,b x) efine by 3.6) an give a concise characterization on their bentness in some special case. The following result is a general characterization on the bentness of f a,b x) efine by 3.6). Proposition 3.14 Assume the notations given as above. Then f a,b x) efine by 3.6) is regular bent if an only if 3 ω Tr 1 bξ j pm 4 ) ω Trn 1 aξ jlx) = 1. x V0 Proof By Lemma.11, f a,b x) is regular bent if an only if p m 4 ) = 1. Note that ωtrn 1 axl )+Tr 1 bx ω Trn 1 ax l ) +Tr 1 bx pm ) 4 = ω Trn 1 ax l ) +Tr 1 b) + aξ l x l) +Tr 1 ω bξ pm ) 4 + ω Trn 1 aξ l x l) Tr 1 b) + aξ 3l x l) +Tr 1 ω = ω Tr 1 b) ω Trn 1 ax) + ω Tr 1 = +ω Tr 1 b) ω Trn 1 aξ l x ) + ω Tr 1 3 ω Tr 1 bξ j pm ) 4 bξ 3 pm ) 4 bξ pm ) 4 ω Trn 1 aξ l x ) bξ 3 pm ) 4 ω 1 Trn aξ jl x ), ω Trn 1 aξ 3l x ) then the result follows. 13

15 On more bent functions from Dillon exponents For given a an b, by Proposition 3.14, it is ifficult to etermine whether f a,b x) efine by 3.6) is a regular bent function. So we consier some special cases of a an b to get a concise characterization for the bentness of f a,b x), an we have the following result. Theorem 3.15 Assume the notations given as above. Let k be a positive integer with k 1or3mo 4),a= aξ k, where a F p m, an 4 l, then f a,bx) efine by 3.6) is regular bent if an only if K m a ) = 1 4I 1sin π Qa) p cos πtr 1 b) p + cos πtr 1 bξ where Qa), I are given in Lemma.7. Proof By Proposition 3.14, f a,b x) is regular bent if an only if Since 4 l, then 3 3 ω Tr 1 bξ j pm 4 ) ω Trn 1 aξ jlx) = 1. x V0 ω Tr 1 bξ j pm 4 ) ω Trn 1 aξ jlx) = x V0 3 p m 4 ) p ω Tr 1 bξ j pm 4 ) ω Trn 1 ax). x V0 Note that a = aξ k, a F pm, k 1or3mo 4), wehave 1 ax) = ω Trn 1 aξ x) = ω Trn 1 aξ 3x) = S 1 a) = S 3 a). ω Trn Hence, by Proposition.8, ω Trn 1 ax) = Ra) I ωqa) ω Qa) ). To sum up, f a,b x) is regular bent if an only if 3 ω Tr 1 bξ j pm 4 ) = 1 K m a ) 4I 1sin where Qa), I are given in Lemma.7. Note that 3 ω Tr 1 bξ j pm ) 4 = cos πtr 1 b) πtr 1 + cos p 4, 3.7) π Qa) p bξ pm 4 p ), 13

16 L. Yu et al. an simplify Eq. 3.7), we finish the proof. From Theorem 3.15, the following result can be obtaine. Corollary 3.16 If there exist a, b) F 3 n F 3 such that f a,b x) efine by 3.6) is a regular bent function, then the number of these regular bent functions is a multiple of 4. Proof Since b F, we get b {α i 3n i 7}, where α is a primitive element in F 3 n. Since ξ is a generator of U, then ξ 3m 4 = α 3n 1 4. Hence, b, bα 3n 1 b, bα 3 3n 1 4 have the same value of cos πtr 1 b) 3 + cos πtr 1 bξ the proof. p m 4 ) 4, 3. This completes Example 3.17 Let l = 4, a = aξ, a F, b F, ξ be a generator of cyclic group U ={x F 3 6 x 33 = 1}, then we have mo 4) an f a,b x) = Tr 6 1 ax144 ) + Tr 1 bx18 ). By using Maple, the number of this binomial regular bent functions is 48. Remark 3.18 From Theorem 3.15, we obtain a new class of binomial p-ary regular bent functions, whose bentness is etermine by Kloosterman sum. It is obvious that the characterization on the bentness of f a,b x) in Theorem 3.15 is concise. 3.. Secon class of p-ary bent functions In the following, we assume s, r are integers with gcs, p m + 1) = 1, an g = gcr, p m + 1). Similar to the secon class of binary bent functions, we iscuss the bentness of the following functions f a,b,r x) = p m g 1 axri+s)pm 1) ) + bx pn 1, 3.8) where a F p n, b F p an f 0) = 0. In what follows, we give a necessary an sufficient conition such that f a,b,r x) efine by 3.8) is regular bent. Base on this characterization, we can obtain several classes of p-ary bent functions from simple approaches. Theorem 3.19 Let the notations be given as above. Then f a,b,r x) efine by 3.8) is regular bent if an only if 1 )) { K m a pm cos πb p = 4I sin πb π Q a) p sin p + ɛ, a C 0 + ; ɛ, otherwise, where ɛ = x g =1, ωtrn 1 axs )+bx pm pm x g =1, ω g 1) axs )+bx pm +

17 On more bent functions from Dillon exponents Proof Note that if x g = 1 an x U, then pm g 1 xri+s) ) = xs ). By Lemma.11, f a,b,r x) is regular bent if an only if pm g 1 ω axri+s) ) +bx pm = 1, which is equivalent to x g =1, ω pm g 1) axs )+bx pm + x g =1, ω Trn 1 axs )+bx pm = ) On the other han, we have ω Trn 1 axs )+bx pm x g =1,,x g =1 = ω Trn 1 axs )+bx pm ω Trn 1 axs )+bx pm =ω b ω Trn 1 axs) +ω b ω Trn 1 aξ s x s) Since gcs, p m + 1) = 1, then x g =1, ω Trn 1 axs )+bx pm. 3.10) ω b ω Trn 1 axs) + ω b ω Trn 1 aξ s x s ) = ω b ω Trn 1 ax) + ω b ω Trn 1 aξ s x) = ω b ω Trn 1 ax) + ω b ω Trn 1 aξξs 1 x)) = ω b S 0 a) + ω b S 1 a). 3.11) From Eqs. 3.9), 3.10), 3.11) an Lemma.7, we complete this proof. Let s 1 be a positive integer, an g = gcr, p m + 1) = 1, s = s 1 r.by3.8), we have f a,b,r x) = p m 1 axri+s 1)p m 1) ) + bx pn ) From Theorem 3.19, we have the following result, which is exact [10, Theorems 10,11]. Corollary 3.0 [10] Let the notations be given as above. 13

18 L. Yu et al. 1. If b = 0 an gcs 1 r, p m + 1) = 1, then f a,0,r x) efine by 3.1) is regular bent if an only if K m a pm ) = 1 ω Trn 1 a).. If b = 0 an gcs 1 r, p m + 1) = 1, then f a,b,r x) efine by 3.1) is regular bent if an only if 1 )) { K m a pm cos πb p = 4I sin πb π Q a) p sin p + ɛ, a C 0 + ; ɛ, otherwise, where ɛ = ω Trn 1 a)+b ω b + 1. Note that if p = 3, g =, then 3m 1 = m 1 1 mo 3). Together with gcs, 3 m + 1) = 1 an b = 0, we have ɛ = ω Trn 1 axs) x =1, = ω Trn 1 axs) x=±1 x=±1 ω 3m 1) axs) + 1 ω Trn 1 axs) + 1 = ω Trn 1 ax) x =1, x=±1 x=±1 ω Trn 1 ax) + 1 = 1. Then, by Theorem 3.19, we have the following result. Corollary 3.1 Assume the notations given as above. Let p = 3, g= an b = 0, then f a,0,r x) efine by 3.8) is regular bent if an only if K m a 3m ) = 0. Note that if p = 3, g = an 3 m 3 mo 4), then 3m 1 = m 1 1 mo 3) an 3m is an even integer. Together with gcs, 3 m + 1) = 1, we have ɛ = x =1, = x=±1 = ω b ω Trn 1 axs )+bx 3m ω Trn 1 axs )+bx 3m x=±1 ω Trn 1 ax) ω b x=±1 x=±1 x =1, ω 3m 1) axs )+bx 3m + 1 ω Trn 1 axs )+bx 3m ω Trn 1 ax) + 1 = Hence, the following result can be erive from Theorem Corollary 3. Assume the notations given as above. Let p = 3, g =, 3 m 3 mo 4) an b = 0, then f a,b,r x) efine by 3.8) is regular bent if an only if K m a 3m ) = 1 1 cos πb 3. 13

19 On more bent functions from Dillon exponents Proof Since ɛ = 1, by Theorem 3.19, we have that f a,b,r x) is regular bent if an only if 1 K m a 3m )) cos πb 3 { 4I sin πb π Q a) = 3 sin 3 + 1, a C 0 + ; 1, otherwise. Note that p = 3, 3 m 3 mo 4), then I is a complex number in Lemma.7. Together with real number 1 K m a 3m )) cos πb 3,wehave 1 K m a 3m )) cos πb 3 = 4I 1sin πb 3 sin π Q a) if an only if Q a) = 0, which contraicts with a C 0 +. That is to say f a,b,r x) can not be bent if a C 0 +. This finishes the proof. Remark 3.3 Some other special cases of Theorem 3.19 can also be iscusse as above. From Theorem 3.19, we not only obtain Corollary 3.0, which is exact Theorems 10, 11 in [10], but we also get some characterizations for the bentness of f a,b,r x) efine by 3.8) from simple approaches see Corollaries 3.1, 3.). 4 Conclusion In this paper, we investigate the bentness of several special classes of p-ary functions of the following form f x) = p m 1 ) ) a i x ipm 1) + Tr o) 1 bx pn 1. We obtain some new binary an p-ary bent functions with ifferent kins of Dillon exponents, which inclue binomials an functions with multiple trace terms. All of these bent functions are etermine by some exponential sums, which have close relations with Kloosterman sums. In aition, Corollaries 3.4, 3.7, 3.10, 3.0 obtaine in this paper are the corresponing results in [10]. Acknowlegments We sincerely thank the anonymous reviewers for their helpful comments an suggestions. The work of H. Liu was supporte by NSFC Grant No ) an self-etermine research funs of CCNU from the colleges basic research an operation of MOE Grant No. CCNU14F01004). The work of D. Zheng was supporte by NSFC Grant No ) an the Natural Science Founation of Hubei Province uner Grant 014CFB537. References 1. Canteaut, A., Charpin, P., Kyureghyan, G.: A new class of monomial bent functions. Finite Fiels Appl. 141), ). Charpin, P., Gong, G.: Hyperbent functions, Kloosterman sums an Dickson polynomials. IEEE Trans. Inf. Theory 954), ) 13

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